Question

16.05 g of methane (CH4) gas is mixed with 101.12 g of oxygen (O2) gas. The mixture is ignited. After a bright flash and a loud bang, some water droplets form on the inside of the reaction vessel.

A) Write the balanced chemical reaction for the combustion of methane .

B) How many moles of water can you make?

C) How many moles of CO2 can you make?

D) Will anything be left over? What?

Yes, there will be oxygen left over.

Yes, there will be methane left over.

No, there will be nothing left over. E) Calculate the theoretical yield of water. F) Calculate the theoretical yield of carbon dioxide.

Answer 1

A)
CH4 + 2 O2 ---> CO2 + 2 H2O
B)

Molar mass of CH4,
MM = 1*MM(C) + 4*MM(H)
= 1*12.01 + 4*1.008
= 16.042 g/mol


mass(CH4)= 16.05 g

number of mol of CH4,
n = mass of CH4/molar mass of CH4
=(16.05 g)/(16.042 g/mol)
= 1 mol

Molar mass of O2 = 32 g/mol


mass(O2)= 101.12 g

number of mol of O2,
n = mass of O2/molar mass of O2
=(101.12 g)/(32 g/mol)
= 3.16 moL

In balanced chemical equation
1 mol of CH4 reacts with 2 mol of O2
for 1.000499 mol of CH4, 2.000997 mol of O2 is required
But we have 3.16 mol of O2

so, CH4 is limiting reagent
we will use CH4 in further calculation

According to balanced equation
mol of H2O formed = (2/1)* moles of CH4
= (2/1)*1.000499
= 2.000997 mol


C)
According to balanced equation
mol of CO2 formed = (1/1)* moles of CH4
= (1/1)*1.000499
= 1.000499 mol
D)
Since CH4 is limiting reagent
O2 will be left over
Answer: yes, there will be oxygen left over
E)
Molar mass of H2O,
MM = 2*MM(H) + 1*MM(O)
= 2*1.008 + 1*16.0
= 18.016 g/mol


mass of H2O = number of mol * molar mass
= 2.001*18.02
= 36.05 g

F)
Molar mass of CO2,
MM = 1*MM(C) + 2*MM(O)
= 1*12.01 + 2*16.0
= 44.01 g/mol



mass of CO2 = number of mol * molar mass
= 1*44.01
= 44.03 g